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JEE Advanced · Chemistry · 14. Hydrocarbons

The synthesis of 3-octyne is achieved by adding a bromoalkane into a mixture of sodium amide and an alkyne. The bromoalkane and alkyne respectively are

  1. A \(\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3\) and \(
    \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CH}\)
  2. B \(\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_3\) and \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CH}\)
  3. C \(\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3\) and \(\mathrm{CH}_3 \mathrm{C} \equiv \equiv \mathrm{CH}\)
  4. D \(\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3\) and \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CH}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3\) and \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CH}\)

Step-by-step Solution

Detailed explanation

\(\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_2\) \(-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_3\) There are two (marked) possibility of halides
Thus, in the given context
\(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CH}+\mathrm{NaNH}_2 \longrightarrow \)
\(\mathrm{CH}_3 \mathrm{CH}_2-\mathrm{C} \equiv \overline{\mathrm{CNa}^{+}}\)
\(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \overline{\mathrm{CNa}} \)
\(\longrightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3 \)
\(\text { 3-octyne } \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3\)
Hydrocarbon
Conceptual, reaction understanding II
From JEE Advanced
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